Difference between revisions of "1959 IMO Problems/Problem 2"
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given (a) <math>A=\sqrt{2}</math>, (b) <math>A=1</math>, (c) <math>A=2</math>, where only non-negative real numbers are admitted for square roots? | given (a) <math>A=\sqrt{2}</math>, (b) <math>A=1</math>, (c) <math>A=2</math>, where only non-negative real numbers are admitted for square roots? | ||
− | == Solution == | + | == Solution 1 == |
The square roots imply that <math>x\ge \frac{1}{2}</math>. | The square roots imply that <math>x\ge \frac{1}{2}</math>. | ||
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− | + | == Solution 2 == | |
+ | Note that the equation can be rewritten to | ||
<cmath>\sqrt{(\sqrt{2x-1}+1)^2} + \sqrt{(\sqrt{2x-1}-1)^2}=A\sqrt{2}</cmath> | <cmath>\sqrt{(\sqrt{2x-1}+1)^2} + \sqrt{(\sqrt{2x-1}-1)^2}=A\sqrt{2}</cmath> | ||
i.e., <math>\sqrt{2x-1}+1 + |\sqrt{2x-1}-1|=A\sqrt{2}</math>. | i.e., <math>\sqrt{2x-1}+1 + |\sqrt{2x-1}-1|=A\sqrt{2}</math>. | ||
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In summary, any <math>x \in \left[1/2, 1\right]</math> is a solution for (a); there is no solution for (b); there is one solution for (c), which is <math>x=\frac{3}{2}</math>. | In summary, any <math>x \in \left[1/2, 1\right]</math> is a solution for (a); there is no solution for (b); there is one solution for (c), which is <math>x=\frac{3}{2}</math>. | ||
− | ~zhaoxiayu | + | ~zhaoxiayu |
+ | |||
+ | |||
{{Alternate solutions}} | {{Alternate solutions}} |
Revision as of 20:18, 6 September 2021
Contents
Problem
For what real values of is
given (a) , (b)
, (c)
, where only non-negative real numbers are admitted for square roots?
Solution 1
The square roots imply that .
Square both sides of the given equation:
Add the first and the last terms to get:
Multiply the middle terms, and use to get:
Since the term inside the square root is a perfect square, and by factoring 2 out, we get
Use the property that
to get
Case I: If , then
, and the equation reduces to
. This is precisely part (a) of the question, for which the valid interval is now
Case II: If , then
and we have
which simplifies to
This tells there that there is no solution for (b), since we must have
For (c), we have , which means that
, so the only solution is
.
~flamewavelight (Expanded)
~phoenixfire (edited)
Solution 2
Note that the equation can be rewritten to
i.e.,
.
Case I: when (i.e.,
), the equation becomes
. For (a), we have
; for (b) we have
but
, thus no solution; for (c) we have
.
Case II: when (i.e.,
), the equation becomes
, which only works for (a)
.
In summary, any is a solution for (a); there is no solution for (b); there is one solution for (c), which is
.
~zhaoxiayu
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1959 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |